You may also like

problem icon

8 Methods for Three by One

This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?

problem icon

Figure of Eight

On a nine-point pegboard a band is stretched over 4 pegs in a "figure of 8" arrangement. How many different "figure of 8" arrangements can be made ?

problem icon

Sine and Cosine for Connected Angles

The length AM can be calculated using trigonometry in two different ways. Create this pair of equivalent calculations for different peg boards, notice a general result, and account for it.


Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Sue, Madras College, arrived at the correct answer using the formula $\frac{1}{2}ab \sin C$. For those not familiar with this formula then do look again at the problem. You may find a more accessible solution if you remember Pythagoras and that:

$26 = 5^2 + 1^2$
$20 = 4^2 + 2^2$
$18 = 3^2 + 3^2$.

This solution was submitted by Sue.

First we change square centimeters into square metres, so $26000$ cm$^2 = 2.6$ m$^2$ , $18000$ cm $^2 = 1.8$ m$^2$ and $20000$ cm$^2 = 2$ m$^2$.

Then we put letters at all points shown on the diagram so the three enclosed triangles are triangle $CC_1C_2$, triangle $BB_1B_2$, and triangle $AA_1A_2$.

Let the length of $AB$ be $c$, length of $AC$ be $b$ and the length of $BC$ be $a$. Because of the squares, $BB_1 = AA_2 = c$, $AA_1 = CC_2 = b$ and $CC_1 = BB_2 = a$.

We know the formula for the area of a triangle $ABC$ is $(1/2)absin C$.

In triangle $CC_1C_2$

$CC_1 = CB = a$
$CC_2 = AC = b$
$\sin(180^{\circ} - C) = \sin C$ (because of the two right angles)

so triangle $CC_1C_2$ and triangle $ABC$ are equal in area.

Similarly we can prove triangle $AA_1A_2$ and triangle $BB_1B_2$ are also equal in area to triangle $ABC$. Therefore the four triangles are equal in area and we only need to work out one of these areas.

Using the formula $(1/2)ab \sin C$

Area of triangle $ABC$ = $1/2(\sqrt{2.6})(\sqrt{1.8})\sin{C}$ m$^2$

Using the cosine rule we can work out that

$\eqalign{ \cos{C} &=& \frac{2.6 + 1.8 - 2}{2\sqrt{2.6}\sqrt{1.8}} \\ \; &=& \frac{1.2}{\sqrt{4.68}} \\ \; &=& \frac{2}{\sqrt{13}}}$

Therefore we know that

$\eqalign{ \sin{C} &=& \frac{\sqrt{13 - 2^{2}}}{\sqrt{13}} \\ \; &=& \frac{3}{\sqrt{13}}}$

Put this in our original equation

$\eqalign{ \mbox{Area of triangle} ABC &=& 1/2(\sqrt{4.68})(\frac{3}{\sqrt{13}}) m^2 \\ \; &=& 1/2(0.6\sqrt{13})(\frac{3}{\sqrt{13}}) m^2 \\ \; &=& 0.9 m^2}$

Therefore the area of the hexagon is

(0.9 m$^2$ \times4) + 2.6 m$^2$ + 1.8 m$^2$+ 2 m$^2$ = 3.6 m$^2$ + 6.4 m$^2$
  = 10 m$^2$
  = 100000 cm$^2$